Question

Question: At room temperature, the edge length of the cubic unit cell in elemental silicon is $\mathbf{5}\mathbf{.}\mathbf{431}\stackrel{\mathbf{\circ }}{\mathbf{A}}$, and the density of silicon at the same temperature is $\mathbf{2}\mathbf{.}\mathbf{328}\mathbf{}\mathit{g}\mathit{c}{\mathit{m}}^{\mathbf{-}\mathbf{3}}$. Each cubic unit cell contains eight silicon atoms. Using only these facts, perform the following operations.

(a) Calculate the volume (in cubic centimeters) of one unit cell.

(b) Calculate the mass (in grams) of silicon present in a unit cell.

(c) Calculate the mass (in grams) of an atom of silicon.

(d) The mass of an atom of silicon is 28.0855 u. Estimate Avogadro’s number to four significant figures.

Short answer

(a) The volume in cubic centimeters, of one unit cell is $160.19\times {10}^{-24}c{m}^3$.

(b) The mass (in grams) of silicon present in a unit cell is $3.729\times {10}^{-22}g$.

(c) The mass (in grams) of an atom of silicon is $4.66\times {10}^{-23}g$.

(d) The estimated Avogadro’s number up to four significant figures is $6.025\times {10}^{23}/mol$.

Step-by-step solution

Subpart (a) The volume of one unit-cell in cubic centimeters.

In cubic systems,

$\alpha =\beta =\gamma =90°$

$a=b=c$

$V=abc\sqrt{1-{\cos }^2\alpha -{\cos }^2\beta -{\cos }^2\gamma +2\cos \alpha \cos \beta \cos \gamma }$

localid="1660652368980" $V=\left(5.431\right)\left(5.431\right)\left(5.431\right)\sqrt{1-{\cos }^2\left(90°\right)-{\cos }^2\left(90°\right)-{\cos }^2\left(90°\right)+2\cos \left(90°\right)\cos \left(90°\right)\cos \left(90°\right)}$

As;

$\cos 90°=0$

$V=160.19\sqrt{1-0}$

$$\begin{gathered} V=160.19\left(1\right) \\ V=160.19\stackrel{\circ }{A^3} \end{gathered}$$

Changing Angstrom to cm;

$$\begin{gathered} 160.19\stackrel{\circ }{A^3}=160.19\times {\left({\text{10}}^{-8}\right)}^{3}c{m}^3 \\ 160.19\stackrel{\circ }{A^3}=160.19\times {\text{10}}^{-24}c{m}^3 \end{gathered}$$

Subpart (b) The mass (in grams) of silicon present in a unit cell.

As;

$density=\frac{mass}{volume}$

Therefore;

$$\begin{gathered} mass=density\times volume \\ mass=2.328{\text{g cm}}^{-3}\times 160.19\times {10}^{-24}c{m}^3 \\ mass=3.729\times {10}^{-22}g \end{gathered}$$

Subpart (c) The mass (in grams) of an atom of silicon.

As each cubic unit cell contains eight silicon atoms;

Therefore, the mass of an atom is;

$\frac{3.729\times {10}^{-22}g}{8}=4.66\times {10}^{-23}g$

Subpart (d) The estimated Avogadro’s number up to four significant figures.

As known;

$density=\frac{M/N_A}{volume}$

Here;

$density=2.328{\text{g cm}}^{-3}$

$\begin{array}{rcl}\text{M}\left({\text{S}}_8\right)& =& 28.0855\times 8\\ & =& 224.684\text{g/mol}\end{array}$

Putting the values in the equation;

$2.328{\text{g cm}}^{-3}=\frac{8\times 224.684\text{g/mol}}{160.19\times {\text{10}}^{-24}c{m}^3\times N_A/mol}$

$$\begin{gathered} N_A/mol=\frac{224.684\text{g/mol}}{2.328{\text{g cm}}^{-3}\times 160.19\times {\text{10}}^{-24}c{m}^3} \\ {N}_A/mol=\frac{224.684}{3.729\times {10}^{-22}}/mol \\ {N}_A/mol=60.25\times {10}^{22}/mol \\ {N}_A/mol=6.025\times {10}^{23}/mol \end{gathered}$$